What is the digit on the units place in the expanded value of 97^275 –

We have to find unit digit of 97^275–32^44
275=3 mod 4
Hence unit digit of 97^275 is same as 7^3=3

44=0 mod 4
Hence unit digit of 32^44 is same as 2^4=6

unit digit of 97^275–32^44= x3-y6=7

The cyclicity of 7 and 2 is 4.That is the last digit repeats after every four powers.

Say for 7
7
9
3
1
again
7
9
3
1
Similarly for 2 it is 2,4,8,6

Now
97^275 implies last digit is 7 ^ 3= 3 (275/4 leaves 3 )

32^44 implies last digit is 6 (44/2 leaves 0 )

Therefore last digit = 3-6 = 13-6= 7 (last digit will not be negative we will have to borrow one )

What is the digit on the units place in the expanded value of 97^275–32^44?

A. 1
B. 3
C. 5
D. 7
E. 9

In 97^275, unit digit will depend on 7^275 and in 32^44 will depend on 2^44.
Unit digit of
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
In general unit digit of 7^(4x)=1
275=4*68+3
Unit digit of 97^275 = 3

Similarily,
Unit digit of
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6
2^5 = 2
In general unit digit of 2^(4x)=6
44=4*11
Unit digit of 32^44 = 6

Unit digit of 97^275-32^44 = 13- 6 = 7 since 3-6 gives negative value and one is borrowed from 10's digit.

IMO D

Think answer is D.

97^275 - 32^44

7^1=7
7^2=9
7^3=3
7^4=1
7^5=7 .. the trend continues 7,9,3,1

2^1=2
2^2=4
2^3=8
2^4=6
2^5=2 .. the trend continues 2,4,8,6

97^275 - 32^44

Effectively 7^275 - 2^44
7^275 will have the same last digit as 7^3
2^44 will have the same last digit as 2^4

7^3-2^4

3 - 6
(Same as 13 - 6 carried over from the tens digfit)
= 7

Posted from my mobile device

(1) All UNIT DIGITS FOLLOW "THE CYCLICITY OF 4" i.e. after every four powers, UNIT DIGIT REMAINS SAME.
So, Divide the all POWERs by 4 and work with the REMAINDER. (IF the REMAINDER is ZERO, take 4 as the remaining number because you have divided the numbers by 4)

A SHORTCUT regarding the DIVISION OF 4: JUST take LAST TWO DIGIT & DIVIDE them by 4.
For example, 275/4 is same as 75/4 (REMAINDER=3).

(2) When you are asked to find out UNIT DIGIT, work with UNIT DIGIT only (CROSS OUT TENS & HUNDREDS).
For example, UNIT DIGIT of (97^275) and UNIT DIGIT of (9^275) are the SAME.

considering cyclic of 4 & unit digit only, the question {what is the unit digit of( 97^3)-(32^4) becomes what is the unit digit of (7^3)-(2^4)?
Here, UNIT DIGIT of 7^3=3 and
UNIT DIGIT of 2^4=6 [NOTE: After dividing 44 by 4, REMAINDER is ZERO, for UNIT DIGIT CYCLICITY PURPOSE we will take 4 as remaining number because remainder must be an integer between 1&4).

NOW, CHECK whether (97^275) IS GREATER THAN (32^44)?
Case 1: if (97^275) IS GREATER THAN (32^44)?, the value of (97^275)- (32^44)=*****************3-************6= ****************7 (because we consider 3 as 13 because first term is GREATER)
Case 1: if (97^275) IS LESS THAN (32^44)?, the value of (97^275)- (32^44)=*****************3-************6=- **********3 (because simply 6 MINUS 3 because SECOND term is GREATER)

CHECKING:
97>32 &
275>44.
SO, (97^275) IS DEFINITELY GREATER THAN(32^44).
SO, ONLY CASE 1 POSSIBLE.

so, UNIT DIGIT=7 (D is the ANSWER)

in no. 97, cyclicity of unit's digit 7 is 7,9,3,1
and 275 /4 gives 3 as a remainder.
in no.32, cyclicity of unit's digit 2 is 2,4,8,6
and 44/4 gives 0 as remainder.

hence the equation is ...3-...6 which will end up to be 7 as the unit's digit.
hence option D

we need to find the unit digit of
97^275 - 32^44

so we will consider unit digits of both
\(7^1\)=7
\(7^2\)=9 (unit digit)
\(7^3\)=3
\(7^4\)=1
\(7^5\)=7
\(7^6\)=9
\(7^7\)=3
\(7^8\)=1
from here we conclude that it follows a cyclic pattern \(7^4\),\(7^8\) = each unit digit = 1
we need to find unit digit of 7^275
so we will write this in terms of \(7^4\) -- (7^4)^68 * \(7^3\) = 1*\(7^3\) = 3 (unit digit of 97^275)

now unit digit of 2^44
we need to find cyclic pattern
by performing the same operation above on 2 we find pattern \(2^4\),\(2^8\) = each unit digit = 6
so we will write 2^44 in terms of \(2^4\)-- (2^4)^11 = 6^11 = 6 (unit digit of 32^44)

the difference of unit digit
97^275 - 32^44

(...........3) - (....6) = 7 (since 97^275 is larger value than 32^44)

correct answer is 7 option D

The questions tests the knowledge of cyclisity.
Here we are given 2 number but our main focus should be on the last digits of both the numbers, that is, 7 and 2.
Cyclisity of 7 is 4 (7, 9, 3, 1), so in \(7^{275}\), when we divide 275 by 4 and get 3 as a remainder. So the number \(7^{275}\) will end with the last digit of \(7^{3}\) and that is 3.

On the similar lines, the Cyclisity of 2 is 4 (2, 4, 8, 6), so in \(2^{44}\), when we divide 44 by 4 and get 0 as a remainder. When the remainder = 0, the number ends in \(2^{cyclisity}\) = \(2^{44}\), that is 6.

Now we have (Number ending with 3) - (Number ending with 6).
Think of a few examples : 13 - 6 = 7 / 23-16 = 7 / 143-56 = 87 (Units digit is 7).

So, the last digit would be 7. Answer is D.

Relevant Theory -

The answer is D. This is the explanation:
Unit digit for 97^275 can be obtained by multiplying the 7 for 275 times using only the unit digit:
7^1 = 7
7^2 = 49 unit is 9
7^3 = 63 unit is 3
7^4 = 3 * 7 = 21 unit is 1
7^5 = 1 * 7 = 7
7^6 = 7 * 7 = 49 unit is 9
7^7 = 9 * 7 = 63 unit is 3
7^8 = 3 * 7 = 21 unit is 1
Meaning that at every multiple of 4 the unit digit is 1
Thus, 272 is multiple of 4
Therefore; 97^272 = unit digit of 1
97^273 = unit digit of 7
97^274 = unit digit of 9
97^275 = unit digit of 3
Thus 97^275 has unit digit of 3

Similarly, 32^44 will go through the same process using 2
2^1 = 2
2^2 = 2 * 2 = 4 is the unit digit
2^3 = 4 * 2 = 8 is the unit digit
2^4 = 8 * 2 = 16, 6 is the unit digit
2^5 = 6 * 2 = 12, 2 is the unit digit
2^6 = 2 * 2 = 4
2^7 = 4 * 2 = 8
2^8 = 8 * 2 = 16, 6 is the unit digit
This means that at every multiple of 4, 6 is the unit digit.
44 is a multiple of 4 meaning that 32^44 has unit digit of 6

Furthermore; 97^275 - 32^44 = 3 - 6 (using only the unit digits)
Thus, borrow 1 to add to 3 it becomes 13 - 6 = 7
Final answer = 7, ans = D.

Posted from my mobile device

97^275 - 32^44

Following 7's cyclicity of 4 -

7^1 = 7 = ending in 7
7^2 = 49 = ending in 9
7^3 = 343 = ending in 3
7^4 = 2401 = ending in 1

So, 97^275 will be like

7^275 since we are dealing only with digit in units place currently.

Hence,

7^275 is of the type 7^3 and hence will end in a 3

Similarly, following 2's cyclicity of 4 -

2^1 = 2 = ending in 2
2^2 = 4 = ending in 4
2^3 = 8 = ending in 8
2^4 = 16 = ending in 6

So, 32^44 will be like

2^44 since we are dealing only with digit in units place currently.

Hence,

2^44 is of the type 2^4 and hence will end in a 6

Now we have 2 numbers each ending in 3 and 6 respectively.

A difference of the 2 numbers will leave a 7 in the units digit of the resulting number.
Example: 153 - 46 = 107

Hence answer is 7 and option D

Cyclicity of digits.

Units digits :-
7^1 = 7
7^2 = 9 (49)
7^3 = 3 (343)
7^4 = 1 (2401)

As we can see, 7^4 has 1 at its units digit. Thus 7^5 will have a 7 at its units digit and the whole cycle will be repeated.

Now, 97 has 7 at its units place and the 7 will dictate the units digit of all its exponents, i.e. 97^1 has 7 at its units digit, 97^2 will have 9(9409) at its units digit and so on.
Thus 97^275 = [97^(272)]*[97^3]
272 = 4*68
Units digit of 97^272 = 1
Units digit of 97^3 = 3

Thus units digit of 97^275 = 3


Similarly,
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6 (16)
2^5 = 2 (32)

So, units digit of 32^44 will be the same as the units digit of 2^44
Which is the same as the units digit of 2^4 = 6.

So the units digit of the given equation = 3 - 6 = 7.
Hence (D)


P.S. since 97^275 is clearly greater than 32^44, the resulting number when 32^44 is subtracted from 97^275 will be positive.

This one is about cyclicity , so you have to consider cyclicity of the unit digits (7 and 2) in order to find the resultant unit digit:

cyclicity units of 7:

7^1=7
7^2=9
7^3=3
7^4=1

cyclicity units of 2:

2^1=2
2^2=4
2^3=8
2^4=6

So, for 97^275:

275=4K+3, so the unit number will have 3 as units

So, for 32^44:

44=4K, so the unit number will have 6 as units

Then, since 97^275 > 32^44, and the units of each are 3 and 6 respectively, the resultant unit digit must be 7.

D is the answer.

Unit digit of 97^275 would be same as 7^275, which would be same as 7^3 (the unit digits of 7^x repeat after x = 4), which is 3.
Similarly, unit digit of 32^44 would be same as 2^44, which would be same as 2^4 (the unit digits of 2^x repeat after x = 4), which is 6.

Now, ____3-____6 will be same as 13-6 = 7 (where "____x" represents some number string ending with x)
Hence, answer should be (D).

Unit digit of \(97^{275}–32^{44}\) = Unit digit of \(7^{275}–2^{44}\)

CONCEPT: The unit digit of any number depends on the unit digit of the number so any digit other than unit digit is irrelevant such as \(97^{275}\) will have same unit digit as \(387^{275}\) or \(7^{275}\)

Now, Unit digit of \(7^{275}–2^{44}\) = Unit digit of \(7^3–2^4\)

CONCEPT: The cyclicity of unit digit of 7 and 2 is 4 i.e.e unit digits repeat after every 4th power. therefore Remainder (275/4) = 3 hence unit digit of \(7^275\) = unit digit of \(7^3\)

Unit digit of \(7^3–2^4\) = 3-6 = 7

Answer: Option D

We are looking at the unit digits so we just need to consider the unit digit of each number.
So we need to look at
7^275 - 2^44
Now, unit digits of
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
So, we see after every 4th power the unit digit starts repeating.
We essentially need
7^3 as 275 = 4*68 + 3
Similarly
44 is a multiple of 4 so we will look at unit digit of 2^4
The problem is reduced to 3 - 6.
As 3 is a small number than 6 and is present at the units place, it will take a carry over from the tens place.
So, it becomes 13 - 6 = 7

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.